What is the relationship between force and potential in particle interactions?

[Tim667]

TL;DR

Slight confusion about vectors from a potential

Suppose I have some interaction potential, u(r), between two repelling particles. We will name them particles 1 and 2.

I want to find the force vectors F_12 and F_21. Would I be correct in saying that the x-component of F_12 would be given by -du/dx, y-component -du/dy etc? And to find the components for the other force vector, would this simply be the negative of the first vector?

So F_12 would be given by (-du/dx, -du/dy, -du/dz) and F_21= (du/dx, du/dy, du/dz)?

Thank you

 

[PeroK]

Summary: Slight confusion about vectors from a potential

Suppose I have some potential, u(r), between two repelling particles. We will name them particles 1 and 2.

I want to find the force vectors F_12 and F_21. Would I be correct in saying that the x-component of F_12 would be given by -du/dx, y-component -du/dy etc? And to find the components for the other force vector, would this simply be the negative of the first vector?

So F_12 would be given by (-du/dx, -du/dy, -du/dz) and F_21= (du/dx, du/dy, du/dz)?

Thank you

You need to be careful. Normally with a potential you have a particle moving under the influence of an external force. In this case technically you have a different potential for each particle.

 

[Tim667]

You need to be careful. Normally with a potential you have a particle moving under the influence of an external force. In this case technically you have a different potential for each particle.

I see, I should specify that this is an interaction potential between the two particles

 

[PeroK]

I see, I should specify that this is an interaction potential between the two particles

Okay. Of course. As long as you define $r$ the right way round, what you have looks right. I must be getting tired.

 

[Tim667]

Okay. Of course. As long as you define $r$ the right way round, what you have looks right. I must be getting tired.

That's okay. I think I was confused because potentials are usually scalar functions, and I wasn't sure you could get vectors from them

 

[PeroK]

That's okay. I think I was confused because potentials are usually scalar functions, and I wasn't sure you could get vectors from them

Yes, the gradient takes a scalar function of position and generates a vector function of position.

 

[vanhees71]

An interactian potential usually takes the form $V(\vec{x}_1-\vec{x}_2)$. The force on particle 1 is
$$\vec{F}_{12}=-\vec{\nabla}_1 V(\vec{x}_1-\vec{x}_2)$$
and on particle 2
$$\vec{F}_{21}=-\vec{\nabla}_2 V(\vec{x}_1-\vec{x}_2)=-\vec{F}_{12}.$$
Newton's 3rd Law holds, because the potential only depends on the relative vector $\vec{r}=\vec{x}_1-\vec{x}_2$.